## Calculus: Early Transcendentals 8th Edition

$F'(t)=(e^{t\sin2t})(2t\cos2t+\sin2t)$
$F(t)=e^{t\sin2t}$ Differentiate using the chain rule: $F'(t)=(e^{t\sin2t})(t\sin2t)'=...$ Apply the product rule to find $(t\sin2t)'$: $...=(e^{t\sin2t})[(t)(\sin2t)'+(\sin2t)(t)']=...$ Apply the chain rule one more time to find $(\sin2t)'$: $...=(e^{t\sin2t})[(t)(\cos2t)(2t)'+(\sin2t)(1)]=...$ Simplify: $...=(e^{t\sin2t})(2t\cos2t+\sin2t)$